Find The Function Value If Possible

Finding the value of a function is a fundamental skill in mathematics, bridging algebra and calculus. It's about understanding how a function acts as a machine, taking an input and transforming it into a specific output. Mastering this skill opens doors to understanding more complex mathematical concepts and real-world applications.

Understanding Functions: The Basics

At its core, a function is a relationship between a set of inputs and a set of permissible outputs, with the property that each input is related to exactly one output. Think of it as a vending machine: you put in a certain amount of money (the input), select an item (the function's rule), and get a specific snack (the output).

Formal Definition: A function f from a set A to a set B is a rule that assigns to each element x in A exactly one element y in B. We write f(x) = y.

• x is the input or independent variable.
• y is the output or dependent variable.
• A is the domain of the function (all possible inputs).
• B is the codomain of the function (the set where outputs reside).
• The range of the function is the set of all actual outputs (y values) of the function. The range is a subset of the codomain.

Types of Functions:

• Linear Functions: f(x) = mx + b (straight line)
• Quadratic Functions: f(x) = ax² + bx + c (parabola)
• Polynomial Functions: f(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₁x + a₀
• Rational Functions: f(x) = p(x) / q(x) where p(x) and q(x) are polynomials
• Exponential Functions: f(x) = aˣ
• Logarithmic Functions: f(x) = logₐ(x)
• Trigonometric Functions: f(x) = sin(x), cos(x), tan(x), etc.
• Piecewise Functions: Defined by different formulas on different parts of the domain.

Finding the Function Value: Step-by-Step

The process of finding the function value involves substituting a given input value into the function's formula and simplifying the expression. Here's a breakdown:

1. Identify the Function:

Make sure you clearly understand the function you're working with. What is the rule that defines the relationship between input and output? For example:

• f(x) = 3x + 2
• g(t) = t² - 5t + 6
• h(z) = √(z + 4)
• k(x) = { x + 1, if x < 0; x², if x ≥ 0 }

2. Identify the Input Value:

What value are you being asked to substitute into the function? This could be a number, a variable, or even an expression. For example:

• Find f(2)
• Find g(-1)
• Find h(5)
• Find k(-2) and k(3)
• Find f(a + 1)

3. Substitute the Input Value:

Replace every instance of the independent variable in the function's formula with the given input value. Use parentheses to avoid errors, especially when dealing with negative numbers or expressions.

• For f(x) = 3x + 2, to find f(2), substitute x = 2: f(2) = 3(2) + 2
• For g(t) = t² - 5t + 6, to find g(-1), substitute t = -1: g(-1) = (-1)² - 5(-1) + 6
• For h(z) = √(z + 4), to find h(5), substitute z = 5: h(5) = √(5 + 4)
• For k(x) = { x + 1, if x < 0; x², if x ≥ 0 }, to find k(-2), substitute x = -2 (since -2 < 0, use the first rule): k(-2) = -2 + 1; to find k(3), substitute x = 3 (since 3 ≥ 0, use the second rule): k(3) = 3²

4. Simplify the Expression:

Use the order of operations (PEMDAS/BODMAS) to simplify the resulting expression. This often involves performing arithmetic operations, combining like terms, and evaluating exponents and roots.

• f(2) = 3(2) + 2 = 6 + 2 = 8
• g(-1) = (-1)² - 5(-1) + 6 = 1 + 5 + 6 = 12
• h(5) = √(5 + 4) = √9 = 3
• k(-2) = -2 + 1 = -1
• k(3) = 3² = 9

5. State the Function Value:

Clearly state the function value. For example:

• f(2) = 8
• g(-1) = 12
• h(5) = 3
• k(-2) = -1
• k(3) = 9

When is Finding the Function Value Impossible?

Sometimes, finding the function value is impossible because the input value is not in the domain of the function. This can occur for several reasons:

1. Division by Zero:

Rational functions, f(x) = p(x) / q(x), are undefined when the denominator q(x) = 0. You cannot divide by zero.

• Example: f(x) = 1 / (x - 2). Finding f(2) would require dividing by zero: f(2) = 1 / (2 - 2) = 1 / 0, which is undefined. Therefore, x = 2 is not in the domain of f(x).

2. Negative Values Under an Even Root:

Functions involving even roots (square root, fourth root, etc.) are undefined when the radicand (the expression under the root) is negative. In the realm of real numbers, you cannot take the square root of a negative number.

• Example: g(x) = √(x - 3). Finding g(1) would require taking the square root of a negative number: g(1) = √(1 - 3) = √(-2), which is not a real number. Therefore, x = 1 is not in the domain of g(x).

3. Logarithm of a Non-Positive Number:

Logarithmic functions, f(x) = logₐ(x), are only defined for positive values of x. You cannot take the logarithm of zero or a negative number.

• Example: h(x) = ln(x) (natural logarithm). Finding h(0) or h(-1) is impossible because ln(0) and ln(-1) are undefined. Therefore, x = 0 and x = -1 are not in the domain of h(x).

4. Restricted Domains in Piecewise Functions:

Piecewise functions are defined differently on different intervals of the domain. If the input value falls outside the defined interval, the function is undefined at that point.

• Example: k(x) = { x + 1, if x < 0; x², if x ≥ 0 }. If you were asked to evaluate the function at x = -5, the correct piece is x + 1, so k(-5) = -5 + 1 = -4. But if you were asked to evaluate at x = 0, the correct piece is x², so k(0) = 0² = 0. Trying to use the wrong piece for a given x value is undefined.

5. Other Contextual Restrictions:

Sometimes, restrictions are imposed based on the context of the problem. For example, if a function represents the number of items produced, the input might be restricted to non-negative integers. Or if a function models a physical quantity like distance, it might only be valid for a certain range of values.

Domain and Range: A Closer Look

Understanding the domain and range of a function is crucial for determining when finding the function value is possible.

Domain:

The domain is the set of all possible input values (x) for which the function is defined. To find the domain, you need to consider the restrictions mentioned above:

• Avoid division by zero.
• Avoid negative values under even roots.
• Avoid logarithms of non-positive numbers.
• Consider any contextual restrictions.

Example 1: Find the domain of f(x) = √(x - 2).

• The expression under the square root must be non-negative: x - 2 ≥ 0
• Solve for x: x ≥ 2
• The domain is all real numbers x such that x ≥ 2. In interval notation, the domain is [2, ∞).

Example 2: Find the domain of g(x) = 3 / (x + 1).

• The denominator cannot be zero: x + 1 ≠ 0
• Solve for x: x ≠ -1
• The domain is all real numbers x such that x ≠ -1. In interval notation, the domain is (-∞, -1) ∪ (-1, ∞).

Range:

The range is the set of all possible output values (y) that the function can produce. Finding the range can be more challenging than finding the domain, and it often requires understanding the behavior of the function.

• For linear functions, the range is typically all real numbers (unless there are contextual restrictions).
• For quadratic functions, the range is determined by the vertex of the parabola.
• For exponential functions, the range is typically positive real numbers.
• For functions with restricted domains, the range may also be restricted.

Example 1: Find the range of f(x) = x².

• Since x² is always non-negative, the output is always greater than or equal to zero.
• The range is all real numbers y such that y ≥ 0. In interval notation, the range is [0, ∞).

Example 2: Find the range of g(x) = -x² + 5.

• The function is a parabola opening downwards (due to the negative coefficient of x²).
• The vertex of the parabola is at (0, 5), which is the maximum point.
• The range is all real numbers y such that y ≤ 5. In interval notation, the range is (-∞, 5].

Examples and Practice Problems

Let's work through some examples to solidify your understanding.

Example 1:

Given f(x) = 2x³ - x + 4, find f(-2).

• Substitute x = -2: f(-2) = 2(-2)³ - (-2) + 4
• Simplify: f(-2) = 2(-8) + 2 + 4 = -16 + 2 + 4 = -10
• Therefore, f(-2) = -10.

Example 2:

Given g(x) = (x + 3) / (x - 1), find g(1).

• Substitute x = 1: g(1) = (1 + 3) / (1 - 1) = 4 / 0
• Since we have division by zero, g(1) is undefined. Therefore, x = 1 is not in the domain.

Example 3:

Given h(x) = √(9 - x²), find h(5).

• Substitute x = 5: h(5) = √(9 - 5²) = √(9 - 25) = √(-16)
• Since we have the square root of a negative number, h(5) is not a real number and is undefined. Therefore, x = 5 is not in the domain.

Example 4:

Given k(x) = { x² + 1, if x < 1; 3x - 2, if x ≥ 1 }, find k(0) and k(3).

• For k(0), since 0 < 1, use the first rule: k(0) = 0² + 1 = 1
• For k(3), since 3 ≥ 1, use the second rule: k(3) = 3(3) - 2 = 9 - 2 = 7

Practice Problems:

1. If f(x) = 4x - 7, find f(3) and f(-1).
2. If g(x) = x² + 2x - 3, find g(0), g(2), and g(-3).
3. If h(x) = 5 / (x + 2), find h(3) and explain why h(-2) is undefined.
4. If k(x) = √(x + 5), find k(4) and explain why k(-6) is undefined.
5. If m(x) = { 2x, if x ≤ 0; x + 3, if x > 0 }, find m(-4) and m(2).

Function Composition

Beyond simply finding the value of a function at a specific point, understanding function composition is a powerful tool. Function composition involves applying one function to the result of another function.

Definition: Given two functions f(x) and g(x), the composition of f with g, denoted f(g(x)), is defined as applying the function g to x first, and then applying the function f to the result.

How to Find Function Composition:

1. Identify the Inner and Outer Functions: In f(g(x)), g(x) is the inner function and f(x) is the outer function.

2. Evaluate the Inner Function: Find the value of g(x) for the given input x.

3. Substitute into the Outer Function: Substitute the result from step 2 into the outer function f(x). Replace every instance of x in f(x) with the expression g(x).

4. Simplify: Simplify the resulting expression.

Example 1:

Let f(x) = x² + 1 and g(x) = 2x - 3. Find f(g(x)) and g(f(x)).

• f(g(x)) = f(2x - 3) = (2x - 3)² + 1 = (4x² - 12x + 9) + 1 = 4x² - 12x + 10
• g(f(x)) = g(x² + 1) = 2(x² + 1) - 3 = 2x² + 2 - 3 = 2x² - 1

Notice that f(g(x)) and g(f(x)) are generally not equal. The order of composition matters.

Example 2:

Let h(x) = √(x) and k(x) = x + 2. Find h(k(x)) and its domain.

• h(k(x)) = h(x + 2) = √(x + 2)

• To find the domain of h(k(x)) = √(x + 2), we need to ensure that the expression under the square root is non-negative: x + 2 ≥ 0.

• Solving for x, we get x ≥ -2. Therefore, the domain of h(k(x)) is [-2, ∞).

Example 3:

Let f(x) = 1/x and g(x) = x - 1. Find f(g(x)) and its domain.

• f(g(x)) = f(x - 1) = 1 / (x - 1)

• To find the domain of f(g(x)) = 1 / (x - 1), we need to ensure that the denominator is not zero: x - 1 ≠ 0.

• Solving for x, we get x ≠ 1. Therefore, the domain of f(g(x)) is (-∞, 1) ∪ (1, ∞).

Important Considerations for Domain in Function Composition:

When determining the domain of f(g(x)), you must consider two things:

1. The domain of the inner function, g(x): The input x must be in the domain of g(x) for the composition to be defined.

2. The domain of the outer function, f(x), after g(x) has been applied: The output of g(x) must be in the domain of f(x).

Example: Let f(x) = √(x) and g(x) = x² - 4. Find f(g(x)) and its domain.

• f(g(x)) = f(x² - 4) = √(x² - 4)

• To find the domain, we need to ensure that x² - 4 ≥ 0. This means x² ≥ 4. Taking the square root of both sides (and remembering to consider both positive and negative roots), we get |x| ≥ 2. This implies x ≤ -2 or x ≥ 2.

• Therefore, the domain of f(g(x)) is (-∞, -2] ∪ [2, ∞).

Conclusion

Finding the value of a function is a fundamental skill with far-reaching applications. By understanding the definition of a function, the importance of domain and range, and the techniques for substitution and simplification, you can confidently tackle a wide variety of problems. Recognizing when a function value is undefined is just as important as finding it when it exists. Mastering function composition further expands your mathematical toolkit, allowing you to analyze more complex relationships between variables. Practice is key to developing fluency and intuition in this area. As you continue your mathematical journey, the concepts and skills you've learned here will serve as a solid foundation for more advanced topics.